Decomposition Algorithm for a Nonlinear Three-Index Transportation Problem

Abstract

A three-index transportation problem is considered, where the indices stand for supplier, consumer, and type of product. A quite broad statement is allowed, including additional points of production, consumption, and nonlinear transportation costs therein. The solution method is a further development of a universal approach proposed earlier by the authors, which is based on the decomposition of the original problem into a sequence of two-dimensional problems with the recalculation of the objective function coefficients. This paper demonstrates the customization of the method to a nonlinear three-index problem.

1. Introduction

The cost of minimizing a transportation problem is one of the key issues in operation research [1]. It was formulated by F.L. Hitchcock [2] and initially developed by L.V. Kantorovich [3,4,5], T.C. Koopmans [6], D.R. Fulkerson [7], M. Klein [8], and A.L. Lurie [9]. Known algorithms for solving transportation problems are based on the plan improvement method in linear programming [10,11,12,13,14,15,16]. However, extension to broader condition and setting formulations is accompanied by difficulties.

A multi-index transportation problem was considered by T. Motzkin [17] and (under the name of multidimensional transportation problem) by A. Corban [18]. Solving transport-type problems with more than two indices is a complex problem [19]. In these cases, intricate, labor-intensive, specially designed approaches are used. As a rule, new methods are unique for the problem and cannot be applied to a broader class or other types of problems. For instance, a multi-index problem is treated in [20] in a case with three indices with clear extension to more indices, but this technique cannot be used in a nonlinear case. In [21,22], a problem of finding multi-product minimum-cost flows in tree-like networks is considered. A solution is based on the reduction to a set of single-product problems and is limited to networks without cycles. In [23], a four-dimensional linear problem is considered with evolutionary techniques, which do not guarantee convergence. A review [24] refers several approaches that combat the complexities of a multi-index statement via stochastic treatment. An attempt for these complexities to rely on an artificial intelligence approach is made in [25]. Obvious drawbacks here are uninterpretability of the solution and the absence of guarantee. Nonlinearity is one more known and intensively studied extension of the transportation problem [26,27]. Various types of nonlinearity can be introduced [19,28], including capacitated transportation [29] and quadratic costs [30]. Solution algorithms also vary [1] and include using fuzzy variables [31], global optimization for a special type of problem [32], applying axial constraints [33], and genetic algorithms [34,35]. The same picture is shown here: most of the methods are especially designed for a narrow problem statement; the rest use approximations and do not guarantee an optimal solution.

Dimensionality reduction is a topical issue and is used in many problems of various classes [36,37]. In [38], an attempt was made to reduce a variety of transportation problem statements to few representatives via reduction. In [39], a decomposition method for solving the classical transportation problem is proposed based on a special dimensionality reduction technique. The technique was primarily developed for optimizing a network problem by splitting it into simpler parts of lower dimensions [40,41]. A set of applications of the method [39] in further works shows its flexibility. In [42], the method is used to solve a transportation problem with additional production and consumption points (with warehouses) for which quadratic penalties are set. In [43], a case of quadratic objective function is considered. In [44], the approach is extended to more general nonlinear dependencies.

The method [39] can be used directly in cases with many indices. In [45,46], this approach is applied to a linear three-index case. In the present paper, we explain the solution for the most general three-index nonlinear case, which envelopes the cases [44,46]. Our goal is to extend further the class of the problem where the approach [39] is applied directly. The solution to the intermediate problem with linking variables in the iterative process, which relate to the case of two products, is given in particular detail. The presented example illustrates the operation of the algorithm.

2. Problem Statement and Solution Method

A multi-product transportation problem with additional production and consumption points is considered. There are m suppliers $(A_1,A_2,\dots ,A_m)$, n consumers $(B_1,B_2,\dots ,B_n)$, and k different products $(C_1,C_2,\dots ,C_k)$. The supplier $A_i$ produces $a_{it}$ units of product $C_t$. The consumer $B_j$ demands $b_{jt}$ units of product $C_t$. The route from supplier $A_i$ to consumer $B_j$ carries $c_{ij}$ product units. Transporting a unit of product $C_t$ from supplier $A_i$ to consumer $B_j$ costs $d_{ijt}$. All values $a_{it}$, $b_{jt}$, $c_{ij}$, and $d_{ijt}$ are given. In addition, within the total supply volume, each supplier has the opportunity to deliver each product from its own warehouse. The cost of delivery from the warehouse depends nonlinearly on the volume of delivery. The situation is similar for consumers. The corresponding nonlinear dependencies are assumed to be differentiable increasing functions with a non-decreasing derivative. It is necessary to determine a transportation plan with a minimum amount of transportation costs.

A nonlinear three-index transport problem can be formulated as the task of finding values $x_{ijt}$, minimizing the cost functional, as follows:

$$L\left(x\right)=\sum _{i=1}^m\sum _{t=1}^k{f}_{it}\left(y_{it}\right)+\sum _{j=1}^n\sum _{t=1}^k{g}_{jt}\left(w_{jt}\right)+\sum _{i=1}^m\sum _{j=1}^n{h}_{ij}\left(z_{ij}\right)+\sum _{i=1}^m\sum _{j=1}^n\sum _{t=1}^k{d}_{ijt}{x}_{ijt},$$

(1)

subject to the following conditions:

$$y_{it}+\sum _{j=1}^n{x}_{ijt}=a_{it},\forall (i,t)\in N_m\times N_k,$$

(2)

$$w_{jt}+\sum _{i=1}^m{x}_{ijt}=b_{jt},\forall (j,t)\in N_n\times N_k,$$

(3)

$$z_{ij}+\sum _{t=1}^k{x}_{ijt}=c_{ij},\forall (i,j)\in N_m\times N_n,$$

(4)

$$x_{ijt}⩾0,\forall (i,j,k)\in N_m\times N_n\times N_k,$$

(5)

where $N_m$, $N_n$, and $N_k$ are the sets of suppliers, consumers, and products, respectively. Dimensions of the problem, m, n, and k, can be any natural numbers.

For the problem (1)–(5) to be solvable, the following conditions must be met:

$$\sum _{i=1}^m{a}_{it}=\sum _{j=1}^n{b}_{jt},\forall t\in N_k,$$

(6)

$$\sum _{t=1}^k{a}_{it}=\sum _{j=1}^n{c}_{ij},\forall i\in N_m,$$

(7)

$$\sum _{t=1}^k{b}_{ji}=\sum _{i=1}^m{c}_{ij},\forall j\in N_n.$$

(8)

To obtain the first pseudo-solution, $(m+n)k+mn$ partial problems are solved. Coefficients of the objective functions of partial problems are equal to that of function (1) divided by 3, and there is one limitation (see [44]). If the union of the optimal solutions of all $(m+n)k+mn$ partial problems is a feasible solution to the original problem (1)–(5), it means that the solution is reached. Of course, this cannot be counted on. Almost certainly, the values of the same variables in different one-dimensional problems are not equal, and the constructed union of partial solutions is a pseudo-solution.

One can see that the value of the objective function of any pseudo-solution is not greater than that of the optimal solution of the original problem. A sequence of pseudo-solutions with a monotonically increasing objective function can be constructed. To do this, partial problems with three constraints each are cyclically solved; there will be $mnk$ such problems. A problem with one common variable $x_{i_0{j}_0{t}_0}$ and three constraints may be written in general form as follows:

$$y_{i_0{t}_0}+\sum _{j=1}^n{x}_{i_{0}j{t}_0}=a_{i_0{t}_0},$$

(9)

$$w_{j_0{t}_0}+\sum _{i=1}^m{x}_{i{j}_0{t}_0}=b_{j_0{t}_0},$$

(10)

$$z_{i_0{j}_0}+\sum _{t=1}^k{x}_{i_0{j}_{0}t}=c_{i_0{j}_0},$$

(11)

$$\begin{array}{cc}\hfill & f_{i_0{t}_0}\left(y_{i_0{t}_0}\right)+g_{j_0{t}_0}\left(w_{j_0{t}_0}\right)+h_{i_0{j}_0}\left(z_{i_0{j}_0}\right)\hfill \\ \hfill +& \sum _{j=1,j\ne j_0}^n{d}_{i_{0}j{t}_0}^1{x}_{i_{0}j{t}_0}+\sum _{i=1,i\ne j_0}^m{d}_{i{j}_0{t}_0}^2{x}_{i{j}_0{t}_0}+\sum _{t=1,t\ne t_0}^k{d}_{i_0{j}_{0}t}^3{x}_{i_0{j}_{0}t}\hfill \\ \hfill +& d_{i_0{j}_0{t}_0}{x}_{i_0{j}_0{t}_0}\to min.\hfill \end{array}$$

(12)

The coefficients of the objective function (12) with superscripts when solving the very first problem with three constraints with $i_0=1$, $j_0=1$, and $t_0=1$ are equal to those in (1), each divided by three. The upper constraints for the variables included in (9)–(11) are the minima from the three right-hand sides (2)–(4), in which these variables are included.

The solution to the problem (9)–(12) is obtained as a result of a cyclic process of solving three problems, each with two constraints: the first (9), (10); the second (9), (11); and the third (10), (11). The objective function in each of the three problems is the part of the objective function (12) that contains the variables that are in the corresponding constraints. Problems with two constraints are solved as follows (for definiteness, let us consider the problem with constraints (9), (10)).

The only common variable in both constraints is $x_{i_0{j}_0{t}_0}$. Consider the relationship between $d_{i_0{j}_0{t}_0}$ and the other coefficients in (12). Let

$${\displaystyle \frac{2}{3}}{d}_{i_0{j}_0{t}_0}⩽\underset{i\ne i_0,j\ne j_0,t\ne t_0}{min}\left\{\begin{array}{cc}{d}_{i_{0}j{t}_0}^1+d_{i{j}_0{t}_0}^2,& d_{i_{0}j{t}_0}^1+g_{j_0{t}_0}\left(1\right),\\ f_{i_0{t}_0}\left(1\right)+d_{j_0{t}_0}^2,& f_{i_0{t}_0}\left(1\right)+g_{j_0{t}_0}\left(1\right)\end{array}\right\}.$$

Then it is obvious that, in the optimal solution to the problem (9), (10), (12), $x_{i_0{j}_0{t}_0}=min\{a_{i_0{t}_0},b_{j_0{t}_0}\}$, after which the problem (9), (10), (12) will split into two one-dimensional problems.

After solving the problem (9), (10), (12), it is necessary to determine the values of $d_{i_0{j}_0{t}_0}^1$ and $d_{i_0{j}_0{t}_0}^2$ so that the union of the optimal solutions of the problems (9), (12) and (10), (12) coincides with the optimal solution of the problem (9), (10), (12). Let the minimum take place, for example, at $d_{i_0{j}^{*}{t}_0}^1+g_{j_0{t}_0}\left(1\right)$. Then we assign $0<d_{i_0{j}_0{t}_0}^1⩽d_{i_0{j}^{*}{t}_0}^1$, $0<d_{i_0{j}_0{t}_0}^2⩽g_{j_0{t}_0}\left(1\right)$, $d_{i_0{j}_0{t}_0}^1+d_{i_0{j}_0{t}_0}^2=d_{i_0{j}_0{t}_0}$.

After this, it is already possible to formulate, for example, a problem with constraints (9), (11). In the objective function based on (12), the variable $x_{i_0{j}_0{t}_0}$ has a coefficient equal to $d_{i_0{j}_0{t}_0}^1+{\displaystyle \frac{1}{3}}{d}_{i_0{j}_0{t}_0}$.

Let it be now

$${\displaystyle \frac{2}{3}}{d}_{i_0{j}_0}>\underset{i\ne i_0,j\ne j_0,t\ne t_0}{min}\left\{d_{i_{0}j{t}_0}^1+d_{i{j}_0{t}_0}^2,d_{i_{0}j{t}_0}^1+g_{j_0{t}_0}\left(1\right),d_{i{j}_0{t}_0}^2+f_{i_0{j}_0}\left(1\right),f_{i_0{j}_0}\left(1\right)+g_{j_0{t}_0}\left(1\right)\right\}.$$

In this case, by eliminating the common variable, two one-dimensional problems are solved. The following quantities are then calculated:

$$\Delta u_0=f_{i_0{j}_0}\left(p\right)-f_{i_0{j}_0}(p-1)+g_{j_0{t}_0}\left(q\right)-g_{j_0{t}_0}(q-1),$$

$$\Delta u_1=f_{i_0{j}_0}\left(p\right)-f_{i_0{j}_0}(p-1)+d_{i^{*}{j}_0{t}_0}^2,$$

$$\Delta u_2=g_{j_0{t}_0}\left(q\right)-g_{i_0{t}_0}(q-1)+d_{i_0{j}^{*}{t}_0}^1,$$

$$\Delta u_3=d_{i_0{j}^{*}{t}_0}^1+d_{i^{*}{j}_0{t}_0}^2,$$

where p and q are integer values of the arguments of the corresponding nonlinear functions in the optimal solutions of the one-dimensional problems under consideration.

If $\underset{0⩽k⩽3}{max}\Delta u_k=\Delta u_3$ and ${\displaystyle \frac{2}{3}}{d}_{i_0{j}_0{t}_0}>\Delta u_3$, then $x_{i_0{j}_0{t}_0}=0$ and the two-dimensional problem is solved. In this case, in the record of the two-dimensional problem (9), (10), (12), it is necessary to define

$$d_{i_0{j}_0{t}_0}^1⩾d_{i_0{j}^{*}{t}_0}^1,d_{i_0{j}_0{t}_0}^2⩾d_{i^{*}{j}_0{t}_0}^2,d_{i_0{j}_0{t}_0}^1+d_{i_0{j}_0{t}_0}^2={\displaystyle \frac{2}{3}}{d}_{i_0{j}_0{t}_0}.$$

If $\underset{0⩽k⩽3}{max}\Delta u_k=\Delta u_3$ and ${\displaystyle \frac{2}{3}}{d}_{i_0{j}_0{t}_0}=\Delta u_3$, then the corresponding two-dimensional problem is also solved, but its solution is not unique. In this case, we have the following:

$$x_{i_0{j}^{*}{t}_0}+x_{i_0{j}_0{t}_0}=x_{i_0{j}^{*}{t}_0}^{*},x_{i^{*}{j}_0{t}_0}+x_{i_0{j}_0{t}_0}=x_{i^{*}{j}_0{t}_0}^{*},$$

and $d_{i_0{j}_0{t}_0}^1=d_{i_0{j}^{*}{t}_0}^1$, $d_{i_0{j}_0{t}_0}^2=d_{i^{*}{j}_0{t}_0}^2$.

Now let ${\displaystyle \frac{2}{3}}{d}_{i_0{j}_0{t}_0}<\Delta u_3$. In this case, for the previously excluded common variable $x_{i_0{j}_0{t}_0}$, we write (preliminarily!) the equality: $x_{i_0{j}_0{t}_0}=min\{x_{i_0{j}^{*}{t}_0},x_{i^{*}{j}_0{t}_0}\}$ or up to the maximum based on the constraint (11). If the increase of $x_{i_0{j}_0{t}_0}$ has reached the maximum, then the solution to the two-dimensional problem is complete and $0⩽d_{i_0{j}_0{t}_0}^1<d_{i_0{j}^{*}{t}_0}^1$, $0⩽d_{i_0{j}_0{t}_0}^2<d_{i^{*}{j}_0{t}_0}^2$, $d_{i_0{j}_0{t}_0}^1+d_{i_0{j}_0{t}_0}^2={\displaystyle \frac{2}{3}}{d}_{i_0{j}_0{t}_0}$. Otherwise, $\Delta u_k$, $k=\overline{0,3}$ are recalculated and the solution process continues.

If $\underset{0⩽k⩽3}{max}\Delta u_k=\Delta u_2$, then, at ${\displaystyle \frac{2}{3}}{d}_{i_0{j}_0{t}_0}⩾\Delta u_2$, the solution to the problem is completed and $d_{i_0{j}_0{t}_0}^1⩾d_{i_0{j}^{*}{t}_0}^1$, $d_{i_0{j}_0{t}_0}^2⩾g_{j_0{t}_0}\left(q\right)-g_{j_0{t}_0}(q-1)$, $d_{i_0{j}_0{t}_0}^1+d_{i_0{j}_0{t}_0}^2={\displaystyle \frac{2}{3}}{d}_{i_0{j}_0{t}_0}$. At ${\displaystyle \frac{2}{3}}{d}_{i_0{j}_0{t}_0}<\Delta u_2$, the value of $x_{i_0{j}_0{t}_0}$ increases, while $x_{i_0{j}^{*}{t}_0}$ and $w_{j_0{t}_0}^{*}$ decrease until the moment when $\Delta u_2$ ceases to be a maximum. Then all $\Delta u_k$ are recalculated and the solution process continues.

If $\underset{0⩽k⩽3}{max}\Delta u_k=\Delta u_1$, then for ${\displaystyle \frac{2}{3}}{d}_{i_0{j}_0{t}_0}⩾\Delta u_1$, the solution to the two-dimensional problem is complete and $d_{i_0{j}_0{t}_0}^1⩾f_{i_0{j}_0}\left(p\right)-f_{i_0{j}_0}(p-1)$, $d_{i_0{j}_0{t}_0}^2⩾d_{i^{*}{j}_0{t}_0}^2$, $d_{i_0{j}_0{t}_0}^1+d_{i_0{j}_0{t}_0}^2={\displaystyle \frac{2}{3}}{d}_{i_0{j}_0{t}_0}$.

When ${\displaystyle \frac{2}{3}}{d}_{i_0{j}_0{t}_0}<\Delta u_1$, the recalculation of variables and the further solution process are similar to the case ${\displaystyle \frac{2}{3}}{d}_{i_0{j}_0{t}_0}<\Delta u_2$.

If $\underset{0⩽k⩽3}{max}\Delta u_k=\Delta u_0$ and ${\displaystyle \frac{2}{3}}{d}_{i_0{j}_0{t}_0}<\Delta u_0$, then $y_{i_0{j}_0}^{*}$ and $w_{i_0{j}_0}^{*}$ decrease, and $x_{i_0{j}_0{t}_0}$ increases until $\Delta u_0$ loses its maximum status. Then all $\Delta u_k$ are recalculated and the solution process continues.

The process is obviously monotonic in increasing the sum of the values of the objective functions of one-dimensional problems, bounded from above and integer, so the limit is reached in a finite number of steps. If an admissible solution to the original problem is obtained in the limit, then it is its optimal solution.

The method has no limitations on the numbers of suppliers, consumers, and types of product. Convergence and issues of degeneration are discussed in [39].

3. A Sample

The example contains two suppliers, two consumers, and two products. The constraints are

$$\begin{array}{cc}{x}_{111}+x_{121}+y_{11}=20,& x_{112}+x_{122}+y_{12}=44,\\ x_{211}+x_{221}+y_{21}=30,& x_{212}+x_{222}+y_{22}=32,\\ x_{111}+x_{211}+w_{11}=22,& x_{112}+x_{212}+w_{12}=40,\\ x_{121}+x_{221}+w_{21}=28,& x_{122}+x_{222}+w_{22}=36,\\ x_{111}+x_{112}+z_{11}=35,& x_{121}+x_{122}+z_{12}=29,\\ x_{211}+x_{212}+z_{21}=27,& x_{221}+x_{222}+z_{22}=35.\end{array}$$

(13)

The objective function is

$$\begin{array}{cc}\hfill & 27x_{111}+36x_{121}+36x_{211}+45x_{221}+36x_{112}+45x_{212}+45x_{221}\hfill \\ \hfill +& 54x_{222}+2y_{11}^3+3y_{12}^2+y_{21}^2+4y_{22}\hfill \\ \hfill +& 2w_{11}^3+3w_{12}^2+w_{21}^2+4w_{22}\hfill \\ \hfill +& 2z_{11}^3+3z_{12}^2+z_{21}^2+4z_{22}\to min.\hfill \end{array}$$

(14)

It is easy to verify that the solvability conditions are satisfied. We will not rely on luck and solve local problems with one constraint each. Let us start immediately with the consideration of triplets of constraints, each with one common variable. Here, we will also limit ourselves to a single solution of three problems, each with two constraints.

The change in the coefficients of the objective function is the only result of solving problems with two constraints. The solution to the original problem is obtained with the end of the changes in the coefficients of the objective function.

The first problem has the following three constraints:

$$\begin{array}{cc}\hfill & x_{111}+x_{121}+y_{11}=20,\hfill \\ \hfill & x_{111}+x_{211}+w_{11}=22,\hfill \\ \hfill & x_{111}+x_{112}+z_{11}=35,\hfill \\ \hfill & 27x_{111}+12x_{121}+12x_{211}+12x_{112}+2y_{11}^3+2w_{11}^3+2z_{11}^3\to min.\hfill \end{array}$$

(15)

The first problem with the first and second constraints is as follows:

$$\begin{array}{cc}\hfill & x_{111}+x_{121}+y_{11}=20,\hfill \\ \hfill & x_{111}+x_{211}+w_{11}=22,\hfill \\ \hfill & 18x_{111}+12x_{121}+12x_{211}+2y_{11}^3+2w_{11}^3\to min.\hfill \end{array}$$

(16)

Since $18>2+2$, $x_{111}$ is temporarily excluded and two problems are solved, each with one constraint, as follows:

$$\begin{array}{cc}\hfill & x_{121}+y_{11}=20,\hfill \\ \hfill & 12x_{121}+2y_{11}^3\to min,\hfill \end{array}$$

(17)

that is,

$$\begin{array}{cc}\hfill & 2y_{11}^3+12(20-y_{11})\to min,\hfill \\ \hfill & y_{11}^2=2.\hfill \end{array}$$

(18)

The integer minimum is achieved when $y_{11}=1$, $x_{121}=19$.

The second problem is

$$\begin{array}{cc}\hfill & x_{211}+w_{11}=22,\hfill \\ \hfill & 12x_{211}+2w_{11}^3\to min,\hfill \end{array}$$

(19)

that is,

$$\begin{array}{cc}\hfill & 2w_{11}^3+12(22-w_{11})\to min,\hfill \\ \hfill & w_{11}^2=2.\hfill \end{array}$$

(20)

The integer minimum is achieved when $w_{11}=1$, $w_{211}=21$.

$$\begin{array}{cc}& \Delta u_0=2+2,\hfill \\ & \Delta u_1=2+12,\hfill \\ & \Delta u_2=2+12,\hfill \\ & \Delta u_3=12+12=24.\hfill \end{array}$$

Since $18<24$, $x_{111}=19$, $x_{121}=0$, $y_{11}=1$, $x_{211}=2$, $w_{11}=1$.

$$\begin{array}{cc}& \Delta u_0=2+2,\hfill \\ & \Delta u_1=2+12,\hfill \\ & \Delta u_2=2+0,\hfill \\ & \Delta u_3=0+12.\hfill \end{array}$$

Since $18>14$, the two-dimensional problem is solved. $d_{111}^1=8$, $d_{111}^2=10$.

Combining optimal solutions of two one-dimensional problems gives

$$\begin{array}{cc}\hfill & x_{111}+x_{121}+y_{11}=20,\hfill \\ \hfill & 9x_{111}+12x_{121}+2y_{11}^3\to min,\hfill \end{array}$$

(21)

and

$$\begin{array}{cc}\hfill & x_{111}+x_{211}+w_{11}=22,\hfill \\ \hfill & 9x_{111}+12x_{211}+2w_{11}^3\to min.\hfill \end{array}$$

(22)

This coincides with the optimal solution of the original problem with two constraints.

The second problem has the following two constraints:

$$\begin{array}{cc}\hfill & x_{111}+x_{121}+y_{11}=20,\hfill \\ \hfill & x_{111}+x_{112}+z_{11}=35,\hfill \\ \hfill & 17x_{111}+12x_{121}+12x_{112}+2y_{11}^3+2z_{11}^3\to min.\hfill \end{array}$$

(23)

Again, we have $17>2+2$, so two one-dimensional problems are solved without $x_{111}$.

The first one has already been solved; let us solve the second one, as follows:

$$\begin{array}{cc}\hfill & x_{112}+z_{11}=35,\hfill \\ \hfill & 12x_{112}+2z_{11}^3\to min,\hfill \end{array}$$

(24)

that is

$$\begin{array}{cc}\hfill & 2z_{11}^3+12(35-z_{11})\to min,\hfill \\ \hfill & z_{11}^2=2.\hfill \end{array}$$

(25)

The integer minimum is achieved at $z_{11}=1$, $x_{112}=34$.

$$\begin{array}{cc}& \Delta u_0=2+2,\hfill \\ & \Delta u_1=2+12,\hfill \\ & \Delta u_2=2+12,\hfill \\ & \Delta u_3=12+12=24.\hfill \end{array}$$

Since $17<24$, then $x_{111}=19$, $x_{121}=0$, $y_{11}=1$, $x_{112}=15$, $z_{11}=1$.

$$\begin{array}{cc}& \Delta u_0=2+2,\hfill \\ & \Delta u_1=2+12,\hfill \\ & \Delta u_2=2+0,\hfill \\ & \Delta u_3=0+12.\hfill \end{array}$$

Here, $17>14$, so the two-dimensional problem is solved and $d_{111}^1=0$, $d_{111}^3=11$.

The third two-dimensional problem is written as

$$\begin{array}{cc}\hfill & x_{111}+x_{211}+w_{11}=22,\hfill \\ \hfill & x_{111}+x_{112}+z_{11}=35,\hfill \\ \hfill & 21x_{111}+12x_{211}+12x_{112}+2w_{11}^3+2z_{11}^3\to min.\hfill \end{array}$$

(26)

Here, again $21>2+2$, so $x_{111}$ is temporarily excluded and two problems are solved, each with one constraint, as follows:

$$\begin{array}{cc}\hfill & x_{211}+w_{11}=22,\hfill \\ \hfill & 12x_{211}+2w_{11}^3\to min.\hfill \end{array}$$

(27)

This problem has already been solved before; we obtained $w_{11}=1$, $w_{211}=21$.

The second one-dimensional problem is

$$\begin{array}{cc}\hfill & x_{112}+z_{11}=35,\hfill \\ \hfill & 12x_{112}+2z_{11}^3\to min.\hfill \end{array}$$

(28)

This problem has already been solved, as follows: $z_{11}=1$, $x_{112}=34$.

$$\begin{array}{cc}& \Delta u_0=2+2,\hfill \\ & \Delta u_1=2+12,\hfill \\ & \Delta u_2=2+12,\hfill \\ & \Delta u_3=12+12=24.\hfill \end{array}$$

We see that $21<24$; therefore, $x_{111}=20$, $x_{211}=1$, $w_{11}=1$, $x_{112}=14$. Since $x_{111}$ has reached its maximum, and neither $x_{211}$ nor $x_{112}$ has become zero, we repeat the calculations. This will lead to the conditions that have already been fulfilled; i.e., the two-dimensional problem is solved, as follows: $d_{111}^2=10.5$, $d_{111}^3=10.5$.

The second problem with three constraints is

$$\begin{array}{cc}\hfill & x_{121}+x_{111}+y_{11}=20,\hfill \\ \hfill & x_{121}+x_{122}+z_{12}=29,\hfill \\ \hfill & x_{121}+x_{221}+w_{21}=28,\hfill \\ \hfill & 36x_{121}+6x_{111}+15x_{122}+15x_{221}+2y_{11}^3+3z_{12}^2+w_{21}^2\to min.\hfill \end{array}$$

(29)

In this problem, the coefficient of $x_{111}$ in the objective function is the same as it was obtained by solving the previous problems with two constraints and writing them as a set of problems with one constraint each. The second problem with three constraints generates three new problems with two constraints. The result of their solution is three new coefficients in the objective functions for the variable $x_{121}$. To solve the example, a total of five cycles of forming eight problems with three constraints and solving the corresponding problems with two constraints are required. Below are the one-dimensional problems, the union of the optimal solutions of which is the optimal solution of the example. Some of these problems have more than one solution. Uniqueness is achieved by matching the optimal solutions with different constraints, as follows:

$$\begin{array}{cc}\hfill & x_{111}+x_{121}+y_{11}=20,\hfill \\ \hfill & 3x_{111}+2x_{121}+2y_{11}^3\to min,\hfill \\ \hfill & x_{111}=0,x_{121}=20,y_{11}=0.\hfill \end{array}$$

(30)

$$\begin{array}{cc}\hfill & x_{111}+x_{112}+z_{11}=35,\hfill \\ \hfill & 10x_{111}+2x_{112}+2z_{11}^3\to min,\hfill \\ \hfill & x_{111}=0,x_{112}=35,z_{11}=0.\hfill \end{array}$$

(31)

$$\begin{array}{cc}\hfill & x_{111}+x_{211}+w_{11}=22,\hfill \\ \hfill & 14x_{111}+6x_{211}+2w_{11}^3\to min,\hfill \\ \hfill & x_{111}=0,x_{211}=21,w_{11}=1.\hfill \end{array}$$

(32)

$$\begin{array}{cc}\hfill & x_{121}+x_{221}+w_{21}=28,\hfill \\ \hfill & 15x_{121}+20.5x_{221}+w{21}^2\to min,\hfill \\ \hfill & x_{121}=20,x_{221}=0,w_{21}=8.\hfill \end{array}$$

(33)

$$\begin{array}{cc}\hfill & x_{121}+x_{122}+z_{12}=29,\hfill \\ \hfill & 19x_{121}+20.5x_{122}+3z_{12}^2\to min,\hfill \\ \hfill & x_{121}=20,x_{122}=6,z_{12}=3.\hfill \end{array}$$

(34)

$$\begin{array}{cc}\hfill & x_{112}+x_{122}+y_{12}=44,\hfill \\ \hfill & 20x_{112}+20.5x_{122}+3y_{12}^2\to min,\hfill \\ \hfill & x_{112}=35,x_{122}=6,y_{12}=3.\hfill \end{array}$$

(35)

$$\begin{array}{cc}\hfill & x_{112}+x_{212}+w_{12}=40,\hfill \\ \hfill & 14x_{112}+28x_{212}+3w_{12}^2\to min,\hfill \\ \hfill & x_{112}=35,x_{212}=0,w_{12}=5.\hfill \end{array}$$

(36)

$$\begin{array}{cc}\hfill & x_{211}+x_{212}+z_{21}=27,\hfill \\ \hfill & 12x_{211}+13x_{212}+z_{21}^2\to min,\hfill \\ \hfill & x_{211}=21,x_{212}=0,z_{21}=6.\hfill \end{array}$$

(37)

$$\begin{array}{cc}\hfill & x_{211}+x_{221}+y_{21}=30,\hfill \\ \hfill & 18x_{211}+20.5x_{221}+y_{21}^2\to min,\hfill \\ \hfill & x_{211}=21,x_{221}=0,y_{21}=9.\hfill \end{array}$$

(38)

$$\begin{array}{cc}\hfill & x_{122}+x_{222}+w_{22}=36,\hfill \\ \hfill & 4x_{122}+18x_{222}+4w_{22}\to min,\hfill \\ \hfill & x_{122}=6,x_{222}=0,w_{22}=30.\hfill \end{array}$$

(39)

$$\begin{array}{cc}\hfill & x_{221}+x_{222}+z_{22}=35,\hfill \\ \hfill & 4x_{221}+18x_{222}+4z_{22}\to min,\hfill \\ \hfill & x_{221}=0,x_{222}=0,z_{22}=35.\hfill \end{array}$$

(40)

$$\begin{array}{cc}\hfill & x_{212}+x_{222}+y_{22}=32,\hfill \\ \hfill & 4x_{212}+18x_{222}+4y_{22}\to min,\hfill \\ \hfill & x_{212}=0,x_{222}=0,y_{22}=32.\hfill \end{array}$$

(41)

4. Conclusions

The proposed method for solving a classical transportation problem with additional production and consumption points and nonlinear transportation costs has been successfully transferred to the case of a three-index multi-product transportation problem. Decomposition into a sequence of two-dimensional problems with the recalculation of the coefficients of the objective functions has proven its universality and applicability to more complex formulations. This makes it a promising tool for solving a wide class of optimization problems, including poorly studied and complex cases of multidimensionality, multi-indexity, and nonlinearity. Further research can be aimed at expanding this approach, refining algorithmic complexity, and optimizing computational procedures. In particular, it can be broadened further and applied to the assignment problem with Boolean variables.